\documentclass[a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{fontspec}
\usepackage{amsthm}
\usepackage{xltxtra}
\usepackage{mflogo,texnames}
\usepackage{graphicx}
\usepackage{titlesec}

\setmainfont{Times New Roman}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{lemma}
\setCJKmainfont[BoldFont=SimHei,ItalicFont = SimSun]{SimSun}

\title{\heiti\zihao{2} 习题18.1}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{计算下列曲面积分}
\subsection{$\iint\limits_{\Sigma} x^{2} \mathrm{~d} S$, 其中 $\Sigma$ 是球面 $x^{2}+y^{2}+z^{2}=4$}
\textbf{解}\quad
由对称性:
$$
	\iint\limits_{\Sigma} x^{2} \mathrm{~d} S=\dfrac{\iint\limits_{\Sigma} x^{2}+y^2+z^2 \mathrm{~d} S}{3}=\dfrac{4\iint\limits_{\Sigma} \mathrm{~d} S}{3}=\dfrac{64\pi}{3}
$$

\subsection{$\iint\limits_{\Sigma}|x y z| \mathrm{d} S$, 其中拋物面 $\Sigma: z=x^{2}+y^{2}, 0 \leqslant z \leqslant 1$}
\textbf{解}\quad
$\sqrt{1+z^2_x+z^2_y}=\sqrt{1+4x^2+4y^2}$,令$D:x^2+y^2\leqslant 1,x,y\geqslant 0$.所以
$$
	\begin{aligned}
		\iint\limits_{\Sigma}|x y z| \mathrm{d} S=4\iint\limits_Dxy(x^2+y^2)\sqrt{1+4x^2+4y^2}\mathrm{d}S
	\end{aligned}
$$
极坐标变换:
$$
	\begin{aligned}
		4\iint\limits_Dxy(x^2+y^2)\sqrt{1+4x^2+4y^2}\mathrm{d}S & =4\int_0^{\pi/2}\mathrm{d}\theta\int_0^1\rho^5\sqrt{1+4\rho^2}\cos\theta\sin\theta\mathrm{d}\rho \\
		                                                        & =\dfrac{125\sqrt{5}-1}{420}
	\end{aligned}
$$
\subsection{$\iint\limits_{\Sigma}(x+y+z) \mathrm{d} S$,其中 $\Sigma$ 为上半球面 $x^{2}+y^{2}+z^{2}=a^{2}, z \geqslant 0$}
\textbf{解}\quad
极坐标变换:$\left\{\begin{array}{l}
		x=a\sin\varphi\cos\theta \\
		y=a\sin\varphi\sin\theta \\
		z=a\cos\varphi
	\end{array}\right.$,从而$\sqrt{EG-F^2}=a\sin\varphi$
$$
	\begin{aligned}
		\iint\limits_{\Sigma}(x+y+z) \mathrm{d} S & =a^3\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}(\sin\varphi(\cos\theta+\sin\theta)+\cos\varphi)\sin\varphi\mathrm{d}\varphi \\
		                                          & =a^3\pi
	\end{aligned}
$$
\subsection{$\iint\limits_{\Sigma}(a x+b y+c z+d)^{2} \mathrm{~d} S$, 其中 $\Sigma$ 为球面 $x^{2}+y^{2}+z^{2}=a^{2}$}
\textbf{解}\quad
由对称性,可知
$$
	\begin{aligned}
		\iint\limits_{\Sigma}(a x+b y+c z+d)^{2} \mathrm{~d} S & =(a^2+b^2+c^2)\iint\limits_{\Sigma}x^2\mathrm{d}S+d^2\iint\limits_{\Sigma}\mathrm{d}S \\
		                                                       & =\dfrac{a^2+b^2+c^2}{3}\iint\limits_{\Sigma}(x^2+y^2+z^2)\mathrm{d}S+d^2\iint\limits_{\Sigma}\mathrm{d}S\\
															   & =4\pi R^2\left(\dfrac{a^2+b^2+c^2}{3}R^2+d^2\right)
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma} z \mathrm{~d} S$, 其中圆锥面 $\Sigma: z=\sqrt{x^{2}+y^{2}}$ 上介于平面 $z=1$ 与 $z=2$ 间的部分}
\textbf{解}\quad
$$
	\begin{aligned}
		z_x                  & =\dfrac{x}{\sqrt{x^2+y^2}} \\
		z_y                  & =\dfrac{y}{\sqrt{x^2+y^2}} \\
		\sqrt{1+z^2_x+z^2_y} & =\sqrt{2}
	\end{aligned}
$$
令$D:1\leqslant x^2+y^2\leqslant 2$
$$
	\begin{aligned}
		\iint\limits_{\Sigma} z \mathrm{~d} S & =\iint\limits_D\sqrt{x^2+y^2}\sqrt{2}\mathrm{d}x\mathrm{d}y        \\
		                                      & =\sqrt{2}\int_0^{2\pi}\mathrm{d}\theta\int_1^2\rho^2\mathrm{d}\rho \\
		                                      & =\dfrac{14\sqrt{2}\pi}{3}
	\end{aligned}
$$

\subsection{$\iint\limits_{\Sigma} \dfrac{1}{x^{2}+y^{2}} \mathrm{~d} S$,其中 $\Sigma$ 是柱面 $x^{2}+y^{2}=R^{2}$ 被平面 $z=0, z=h$ 所截取的部分}
\textbf{解}\quad
$$
	\begin{aligned}
		\iint\limits_{\Sigma} \dfrac{1}{x^{2}+y^{2}} \mathrm{~d} S & =\int_0^h\mathrm{d}z\int_0^{2\pi}\dfrac{R}{R^2}\mathrm{d}\theta \\
		                                                           & =\dfrac{2\pi h}{R}
	\end{aligned}
$$
\section{计算曲面积分 $\iint\limits_{\Sigma}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} S$, 其中 $\Sigma$ 是内接于球面 $x^{2}+y^{2}+z^{2}=a^{2}$ 的八面体$|x|+| y|+|z|=a$ 的表面.}
\textbf{解}\quad
由对称性,只需计算第一象限的$8$倍.$\sqrt{1+z^2_x+z^2_y}=\sqrt{3}$.
$$
	\begin{aligned}
		\iint\limits_{D}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} S & =\iint\limits_{D}\left(x^{2}+y^{2}+(a-x-y)^2\right) \mathrm{d} S                    \\
		                                                            & =\iint\limits_{D}\left(x^{2}+y^{2}+(a-x-y)^2\right)\sqrt{3} \mathrm{d} x\mathrm{d}y \\
		                                                            & =\sqrt{3}\int_0^a\mathrm{d}y\int_0^{a-y}2x^2+2y^2+2xy-2ax-2ay+a^2\mathrm{d}x        \\
		                                                            & =\dfrac{\sqrt{3}a^4}{4}
	\end{aligned}
$$
所以
$$
I=2\sqrt{3}a^4
$$
\section{设 $u(x, y, z)$ 为连续函数,它在 $M\left(x_{0}, y_{0}, z_{0}\right)$ 处有连续的二阶偏导数, $\Sigma$ 为以 $M$ 点为 中心,半径为 $R$ 的球面,以及$$T(R)=\dfrac{1}{4 \pi R^{2}} \iint\limits_{\Sigma} u(x, y, z) \mathrm{d} S$$}
\subsection{证明 :$ \lim\limits_{R \rightarrow 0} T(R)=u\left(x_{0}, y_{0}, z_{0}\right)$}
\begin{proof}
	由积分中值定理:
	$$
		\begin{aligned}
			\lim\limits_{R \rightarrow 0} T(R)=u\left(x_{0}, y_{0}, z_{0}\right) & =\lim\limits_{R \rightarrow 0} \dfrac{4\pi R^2 u(\xi,\eta, \varphi)}{4\pi R^2} \\
			                                                                     & =u(x_0,y_0,z_0)
		\end{aligned}
	$$
\end{proof}
\subsection{若 $\left.\left(\dfrac{\partial^{2} u}{\partial x^{2}}+\dfrac{\partial^{2} u}{\partial y^{2}}+\dfrac{\partial^{2} u}{\partial z^{2}}\right)\right|_{\left(x_{0}, y_{0}, z_{0}\right)} \neq 0$, 求当 $R \rightarrow 0$ 时无穷小量 $T(R)-u\left(x_{0}, y_{0},z_{0}\right)$ 的主要部分.}
\textbf{解}\quad
需要计算关于$R$的线性主部.不妨令$(x,y,z)=(x_0+Ri,y_0+Rj,z_0+Rk)$.
有$i^2+j^2+k^2=1$.
$$
	\begin{aligned}
		T'(R)  & = \dfrac{1}{4\pi}\iint\limits_{\Sigma}\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial R}+\dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial R}+\dfrac{\partial u}{\partial z}\dfrac{\partial z}{\partial R}\mathrm{d}S                                              \\
		       & =\dfrac{1}{4\pi}\iint\limits_{\Sigma}\dfrac{\partial u}{\partial x}i+\dfrac{\partial u}{\partial y}j+\dfrac{\partial u}{\partial z}k\mathrm{d}S                                                                                                                                      \\
		       & =\dfrac{1}{4\pi}\int_0^{2\pi}\mathrm{d}\theta\int_0^\pi\left(\dfrac{\partial u}{\partial x}\sin\varphi\cos\theta+\dfrac{\partial u}{\partial y}\sin\varphi\sin\theta+\dfrac{\partial u}{\partial z}\cos\varphi\right)\sin\varphi\mathrm{d}\varphi                                    \\
		       & =0                                                                                                                                                                                                                                                                                   \\
		T''(R) & =\dfrac{1}{4\pi}\iint\limits_{\Sigma}\sum\limits_{cyc}\left(\dfrac{\partial^2 u}{\partial x^2}\dfrac{\partial x}{\partial R}\dfrac{\partial x}{\partial R}+2\dfrac{\partial^2 u}{\partial x\partial y}\dfrac{\partial x}{\partial R}\dfrac{\partial y}{\partial R}\right)\mathrm{d}S \\
		       & =\dfrac{1}{4\pi}\iint\limits_{\Sigma}\sum\limits_{cyc}\left(\dfrac{\partial^2 u}{\partial x^2}i^2+2\dfrac{\partial^2 u}{\partial x\partial y}ij\right)\mathrm{d}S                                                                                                                    \\
		       & =\dfrac{1}{4\pi}\int_0^{2\pi}\mathrm{d}\theta\int_0^\pi\left(\dfrac{\partial^2 u}{\partial x^2}\sin^2\varphi\cos^2\theta+\dfrac{\partial^2 u}{\partial y^2}\sin^2\varphi\sin^2\theta+\dfrac{\partial^2 u}{\partial z^2}\cos^2\varphi\right)\sin\varphi\mathrm{d}\varphi              \\
		       & +\dfrac{2}{4\pi}\int_0^{2\pi}\mathrm{d}\theta\int_0^\pi\left(\dfrac{\partial^2 u}{\partial x\partial y}\sin^2\varphi\sin\theta\cos\theta+\dfrac{\partial^2 u}{\partial y\partial z}\sin\varphi\sin\theta+\dfrac{\partial^2 u}{\partial z\partial x}\cos\varphi\sin\varphi\cos\theta
		\right)\sin\varphi\mathrm{d}\varphi                                                                                                                                                                                                                                                           \\
		       & =\dfrac{1}{4}\left(\dfrac{4\pi}{3}\dfrac{\partial^2 u}{\partial x^2}+\dfrac{4\pi}{3}\dfrac{\partial^2 u}{\partial y^2}+\dfrac{4\pi}{3}\dfrac{\partial^2 u}{\partial z^2}\right)+0                                                                                                    \\
		       & =\dfrac{1}{3}\sum\limits_{cyc}\dfrac{\partial^2 u}{\partial x^2}
	\end{aligned}
$$
将$R=0$带入可得$T'(0)=0$.将$T(R)$泰勒展开,可得
$$
	T(R)=T(R)+T'(R)+\dfrac{T''(R)}{2}+o(R^2)
$$
所以线性主部为
$$
	\dfrac{1}{6}\sum\limits_{cyc}\dfrac{\partial^2 u}{\partial x^2}
$$
\section{求密度为 $\rho(x, y)=z$ 的抛物球面壳 $z=\dfrac{1}{2}\left(x^{2}+y^{2}\right), 0 \leqslant z \leqslant 1$ 的质量和质心坐标.}
\textbf{解}\quad
$\sqrt{1+z^2_x+z^2_y}=\sqrt{1+x^2+y^2}$.质量:
$$
	\begin{aligned}
		\dfrac{1}{2}\iint\limits_{\Sigma}x^2+y^2\mathrm{d}S & =\dfrac{1}{2}\iint\limits_{\Sigma}(x^2+y^2)\sqrt{1+x^2+y^2}\mathrm{d}x\mathrm{d}y              \\
		                                                    & =\dfrac{1}{2}\int_0^{2\pi}\mathrm{d}\theta\int_0^{\sqrt{2}}\rho^3\sqrt{1+\rho^2}\mathrm{d}\rho \\
		                                                    & =\dfrac{(2+12\sqrt{3})\pi}{15}
	\end{aligned}
$$
设中心坐标为$(0,0,z_0)$.
$$
\begin{aligned}
	I&=\dfrac{1}{4}\iint\limits_{\Sigma}(x^2+y^2)^2\sqrt{1+x^2+y^2}\mathrm{d}x\mathrm{d}y\\
	&=\dfrac{1}{4}\int_0^{2\pi}\mathrm{d}\theta\int_0^{\sqrt{2}}\rho^5\sqrt{1+\rho^2}\mathrm{d}\rho\\
	&=\dfrac{66\sqrt{3}-4}{105}\pi
\end{aligned}
$$
从而质心坐标为
$$
\left(0,0,\dfrac{596-45\sqrt{3}}{749}\right)
$$


\end{document}